Moles are one of the most important concepts taught in general chemistry. Many people make out the mole concept to be difficult, but as you will see, it’s just as simple as a metric conversion. Importance of Moles
Moles
First off, let’s talk about what a mole actually is. A mole is not a set quantity like Avogadro’s number for example. Avogadro’s number is a set quantity (6.0221415 × 10^{23) }and never changes. Unlike Avogadro’s number, a mole is a number that differs depending on what you are talking about. A good example is a dozen.
A dozen you have 12, but 12 what? I can have 12 cars or 12 feathers, and the weight differs depending on the context. This is exactly how a mole works. The amount differs depending on the context. One mole of Arsenic is completely different than a mole of Aluminum.
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Alright, enough of the explanations, let’s dig into exactly how you can find moles of a substance.
Mole Conversion
To understand how to do a mole to mole conversion (molar conversion), you must first understand the concepts of grams and moles. There is a relationship between Grams (a measurable unit) and Moles.
The best way to see this relationship is to do a few simple mole problems. From the information provided above, you should NOT be able to do the following problems, they should be showing you how to do conversions. To test your skills, see the Mole Worksheet.
Problem 1 

Problem: A chemist needs to measure five moles of Aluminum for his Aluminum Foil Company. Help him figure out how many grams he should weigh out. Plan: In order to find how many grams he needs to weigh out, we need to calculate the grams for five moles of Aluminum. By looking at the periodic table, we see that Aluminum has an atomic mass of 26.982. This means that Aluminum has 26.982 grams/mol. Thus the equation to find how many grams he needs is:

Problem 2 

Problem: Find the Molar Mass of Ethanol (C_{2}H_{5}OH) Plan: To find the molar mass, we must find the atomic masses and multiply them by their subscript for each element. C_{2} = (2)12.01, H_{5}=(6)1.008, O=16.00 
Now that you have the hang of adding up and finding molar masses, let’s try a stoichiometry problem. A stoichiometry problem is where you begin with a number of moles (or grams) and you use the equation to find another quantity of moles or grams in the equation.
Read more about More:
http://chemistry.bd.psu.edu/jircitano/mole.html
In order to do stoichiometry, you must understand that you can find any quantity of any reactant OR product in the equation by just knowing one quantity! Importance of Moles
This is what makes stoichiometry such a powerful tool in the chemistry world. Again, let’s try a sample problem and show you exactly what is meant by all this garbled text.
Problem 3 

Problem: Glucose is the main product of photosynthesis, and is a simple sugar. How many grams of CO_{2} are produced in the following reaction if you begin with 1.00 grams of glucose (c_{6}H_{12}O_{6)? }The molar mass of glucose is 180.16 grams/mol, and the molar mass of CO_{2} is 44.01 grams/mol (However, you should be able to calculate the molar masses by using the same method as in Problem 2.2)C_{6}H_{12}O_{6(s) + }6O_{2(g) }6CO_{2(g) +}6H_{2}O_{(l)} Plan: First we will need to convert the 1.00 grams of glucose into moles (the molar mass is given) so we can perform stoichiometry. The next step will be to form a ratio between C_{6}H_{12}O_{6 and }CO_{2} so we can find moles CO_{2}. Lastly, we will add up the molar mass of CO_{2} and convert it to grams. Comment: Although we haven’t talked about balancing equations yet, in order to do stoichiometry, your equation must be balanced! This is extremely important! See the Balancing Equations section to learn how to balance an equation. Importance of Moles Also, notice how every unit cancels in the problem. This is a way to double check if you have your units right and to make sure nothing is upside down. 
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